Showing that the omega and the alpha limits are disjoint or have just one common point Let $f:\mathbb R^2\rightarrow \mathbb R^2$ be a $C^1$ function and $x'=f(x)$. Suppose that there are finites points $x_i\in \mathbb R^2$, such that $f(x_i)=0$. Given $y$, such that $f(y)\neq 0$, and the flux through $y$ is not periodic, then the $\omega$-limit and the $\alpha$-limit are disjoint, or both are equal to $\{x_0\}$, and moreover $f(x_0)=0$.
I'm trying to show that if they are not disjoint, then there is one point, by supposing that there is another point in one of them, and then getting an absurd. But I can't get it done, hope that you have some tips to me.
Thanks in advance.
 A: The proof is very long. So I skip the details here.
Part 1, you should prove that $f(x_0)=0$. It is not simple. I need a figure to explain it.
To use the proof of contradiction, suppose that $f(x_0)\neq 0$. The proof is divided into 3 steps. 
In Step 1, to show that there exists a small segment $\Lambda$ traversing the curve $\{\phi(x_0,t),t\in D(x_0)\}$ at the point $x_0$. Furthermore, all the curves (the solutions of the ODE) that pass through $\Lambda$ are from the left hand side to the right hand side, as shown in Fig 1.
In Step 2, you should show there exists two different points $x_1$ and $x_2$ on $\Lambda$, such that $x_1=\phi(y,t_1)$ and $x_2=\phi(y,t_2)$, where $t_2>t_1$, as shown in Fig 2.
In Step 3, define a open set S as shown in Fig 3, such that $\partial S=\{\phi(y,t)|t_1\leq t\leq t_2\}\cup \overline{x_1x_2}$. Then you can show that $\phi(y,t)\in \bar{S}$ for all $t\leq t_2$.
Thus, you get a contradiction that $x_0$ can not be the alpha limit of $y$, if $f(x_0)\neq 0$.
Furthermore, you can prove that, if the intersection of the $\alpha$ and $\omega$ limit sets is not empty, all the points in the union of $\alpha$ and $\omega$ limit sets should satisfy $f(x)=0$.

Part 2, prove that both $\omega$ and $\alpha$ limit sets are $\{x_0\}$. Some more definitions and configurations are needed in my method of the proof.
Step 1, we should define a new space $W=\mathbb{R}^2\cup \{\infty\}$, and the open sets in this space are generated by those sets like $\{x:\|x\|>r_1\}\cup\{\infty\}$ and $\{x:\|x-z_0\|<r_2\}$. Then this topology make this space to be a compact space. Also define $f(\infty)=0$, then $\phi(\infty,t)=\infty$ for all $t\in\mathbb{R}$ can also be regarded as a solution of the ODE. The existence and uniqueness of the solution of the ODE on $W$ is still hold.
Step 2, to show that, in this new space, the $\alpha$ and $\omega$ limit sets are both connected. So they can not only contain the finite isolated equilibriums of the ODE, unless there is only one point in both of the two limit sets.
Then, the proof is completed here.
Or there is another way to prove Part 2. You do not need to define a new space here. Notice that there is only finite equilibriums of the ODE on the plane, denoted by $z_1,\ldots,z_n$, respectively. Then, there exists positive numbers $r_1,\ldots,r_n$ such that the sets $A_i\triangleq\{x:\|x-z_i\|<r_i\}$ have no intersection. And there exists a positive number $R>0$ such that $A_i\subset U$, where $U\triangleq \{x:\|x\|<R\}$. Then $\bar{U}$ is compact. And you can prove that if y have more than one $\omega$ limit points $x_0,\tilde{x_0}\in \{z_1,\ldots,z_n\}$, then y must have other $\omega$ limit point in $\bar{U}\backslash \left\{\cup_{i=1}^n A_i\right\}$. This limit point is not a equilibrium of the ODE, which is a contradiction.
